Question

[Nương Mạnh]

 

Thực hiện phép tính

Answer 1

[Xyz OLM]

\(\frac{\left(x^3+1\right)\left(x^6+1\right)}{x^{24}+1}.\frac{\left(x^{12}+1\right)\left(x^{24}+1\right)}{x^{24}-1}\)

\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{24}-1\right)}\)

\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^{12}-1\right)}\)

\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^6-1\right)}=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^3+1\right)\left(x^3-1\right)}\)

\(=\frac{1}{x^3-1}\)

Answer 2

[Hồ Bích Ngọc]

hi nha bạn 2k7 à